Chapter 7, Problem 7.73P

To determine

(a)

Find the terminal ascent velocity?

Answer to Problem 7.73P

$V=9.12m/s$

Explanation of Solution

Given information:

Diameter of balloon is equal to $4m$

Balloon contains helium at $15°C$

The total weight of balloon is equal to $200N$

The drag force is defined as,

$F_{drag}=C_D\frac{\rho V^{2}A}{2}$

In above equation,

$\rho$ - Density of the fluid

$A$ - Characteristic area

$V$ - Velocity

$C_D$ - Drag co-efficient

The buoyancy force is defined as,

$F_{Bouyancy}=\left(Vol\right)\rho g$

The specific heat capacity $R$ is defined as,

$R=\frac{P}{\rho T}$

$P$ - Pressure

$\rho$ - Density

$T$ - Temperature

Assume, the sea level air will have,

$\begin{array}{l}{\rho }_{air}=1.2kg/m^3\\ \mu =1.8\times {10}^{-5}kg/ms\end{array}$

Assume,

The specific heat capacity of helium is equal to,

$R=2077J/kg°K$

Calculation:

Calculate the density of helium,

$\begin{array}{l}R=\frac{P}{\rho T}\\ 2077J/kg°K=\frac{125000Pa}{{\rho }_{He}\left(288°K\right)}\\ {\rho }_{He}=0.21kg/m^3\end{array}$

For force balance,

$F_{Bouyancy}=W+F_{Drag}$

Where,

$W$ - Weight

Substitute,

$\left(\frac{4}{3}\pi {\left(\frac{D}{2}\right)}^3\right)\left({\rho }_{air}-{\rho }_{He}\right)g=W+C_D\frac{\rho V^{2}A}{2}$

Assume, $C_D=0.2$

Substitute for known values,

$\left(\frac{4}{3}\pi {\left(2m\right)}^3\right)\left(1.2kg/m^3-0.21kg/m^3\right)\left(9.81m/s^2\right)=200N+\left(0.2\right)\frac{\left(1.2kg/m^3\right)\left(\pi {\left(2m\right)}^2\right)V^2}{2}$

Solve for velocity $V$,

$V=9.12m/s$

Conclusion:

The terminal ascent velocity is equal to $V=9.12m/s$.

To determine

(b)

Find the altitude when balloon comes to rest?

Answer to Problem 7.73P

$Z\approx 4000m$

Explanation of Solution

Given information:

Diameter of balloon is equal to $4m$

Balloon contains helium at $15°C$

The total weight of balloon is equal to $200N$

According to sub-part a,

We have found,

${\rho }_{He}=0.21kg/m^3$

The buoyancy force is defined as,

$F_{Bouyancy}=\left(Vol\right)\rho g$

Calculation:

If the balloon comes to rest,

The drag will be equal to zero,

Therefore, the force balance equation can be written as,

$F_{Bouyancy}=W$

Substitute,

$\left(\frac{4}{3}\pi {\left(\frac{D}{2}\right)}^3\right)\left({\rho }_{air}-{\rho }_{He}\right)g=W$

Substitute for known values,

$\left(\frac{4}{3}\pi {\left(2m\right)}^3\right)\left({\rho }_{air}-0.21kg/m^3\right)\left(9.81m/s^2\right)=200N$

Solve for ${\rho }_{air}$

${\rho }_{air}=0.8184kg/m^3$

Therefore to find the relevant altitude,

According to table $A.6$ properties of the standard atmosphere,

We can evaluate,

$Z\approx 4000m$

Conclusion:

The final standard altitude is equal to $Z\approx 4000m$.

To determine

(c)

Find the minimum diameter of balloon when it begins to rise?

Answer to Problem 7.73P

$D>3.4m$

Explanation of Solution

Given information:

Diameter of balloon is equal to $4m$

Balloon contains helium at $15°C$

The total weight of balloon is equal to $200N$

According to sub-part a,

We have found,

${\rho }_{He}=0.21kg/m^3$

The buoyancy force is defined as,

$F_{Bouyancy}=\left(Vol\right)\rho g$

Calculation:

If the balloon begins to rise at sea level,

The buoyancy force will be larger than the weight

Therefore,

$F_{Bouyancy}>W$

Substitute,

$\left(\frac{4}{3}\pi {\left(\frac{D}{2}\right)}^3\right)\left({\rho }_{air}-{\rho }_{He}\right)g>W$

Substitute for known values,

$\left(\frac{4}{3}\pi {\left(\frac{D}{2}\right)}^3\right)\left(1.2kg/m^3-0.21kg/m^3\right)\left(9.81m/s^2\right)>200N$

Solve for $D$

$D>3.4m$

Conclusion:

The diameter of the balloon needs to be larger than $3.4m$ to slowly rise at sea level.